The scalar field $f(x, y) = 2\sin(3x) - \cos(2y)$ has a critical point at $\left( \dfrac{5\pi}{6}, \dfrac{3\pi}{2} \right)$. How does the second partial derivative test classify this critical point? Choose 1 answer: Choose 1 answer: (Choice A) A Local maximum (Choice B) B Local minimum (Choice C) C Saddle point (Choice D) D The test is inconclusive
Solution: The second partial derivative test uses the quantity below, evaluated at the critical point we wish to classify. $H = f_{xx}f_{yy} - f_{xy}f_{yx}$ $H < 0$ implies a saddle point. $H > 0$ and $f_{xx} > 0$ implies a local minimum. $H > 0$ and $f_{xx} < 0$ implies a local minimum. $H = 0$ means the test is inconclusive. Let's calculate $H$. First we need all the regular partial derivatives. $\begin{aligned} f_x &= 6\cos(3x) \\ \\ f_y &= 2\sin(2y) \end{aligned}$ Now we can find all the second order partial derivatives. $\begin{aligned} f_{xx} &= -18\sin(3x) = -18 \\ \\ f_{yx} &= 0 \\ \\ f_{xy} &= 0 \\ \\ f_{yy} &= 4\cos(2y) = -4 \end{aligned}$ Therefore, $H = (-18)(-4) - (0)(0) = 72$. Because $H$ is positive, the critical point is either a local maximum or a local minimum. To find which, we can notice that $f_{xx} < 0$, so the critical point is a local maximum.